Question

When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2, how do you determine the remainder when the polynomial is divided by (x+2)(x-1)?

Answer 1

The remainder when dividing `f(x)` by `(x+2)(x-1)` is `(7x-5)`

Explanation:

Note that `(x+2)(x-1) = x^2+x-2` is quadratic, so if we divide another polynomial by it then the remainder will be `0`, a non-zero constant or a linear polynomial. Any remainder of greater degree could be divided further.

Suppose `f(x) = (x+2)(x-1)k(x) + rx + s`

Then:

`f(x)/(x+2) = (x-1)k(x) + (rx+s)/(x+2)`

`=(x-1)k(x) + (rx + 2r -2r + s)/(x+2)`

`=(x-1)k(x) + r + (s-2r)/(x+2)`

Hence [1]: `s-2r = -19`

`f(x)/(x-1) = (x+2)k(x) + (rx+s)/(x-1)`

`=(x+2)k(x) + (rx-r+r+s)/(x-1)`

`=(x+2)k(x)+r+(r+s)/(x-1)`

Hence [2]: `r+s = 2`

Subtract [2] from [1] to get `-3r = -21`, hence `r = 7`

Then `s = 2 - r = -5`

So the remainder when dividing `f(x)` by `(x+2)(x-1)` is `7x-5`

Answer 2

The remainder is `7x-5`.

Explanation:

Call the polynomial `P(x)`.

Because `P(x)` divided by `x+2` leaves a remainder of `-19`, we know that:
`P(x) = (x+2)Q_1(x) - 19`. (Division Algorithm)

Because `P(x)` divided by `x-1` leaves a remainder of `2`, we know that `P(1) = 2` (Remainder Theorem)

From these two facts we get:

`P(1) = (1+2)Q_1(1)-19 = 2`

So `3Q_1(1) = 21`

and `Q_1(1) = 7`

Applying the Division Algorithm and the Remainder Theorem to `Q_1(x)` we learn that:

`Q_1(x) = (x-1)Q_2(x)+7`

Subtitutuing in `P(x) = (x+2)Q_1(x) - 19`, we get:

`P(x) = (x+2)[(x-1)Q_2(x)+7] - 19`

`= (x+2)(x-1)Q_2(x) + 7(x+2) - 19`

`= (x+2)(x-1)Q_2(x) + 7x+ 14 - 19`

`= (x+2)(x-1)Q_2(x) + [7x - 5]`

The remainder is `7x-5`.

Answer 3

`7x-5.`

Explanation:

Let the Poly. in question be `p(x).`

We know that, the Degree of the Remainder Poly. ** is strictly**

less than that of the Divisor Poly.

So, when `p(x)` is diveded by a Quadr. Poly. `(x+2)(x-1)`, the

Remainder Ploly. must have a degree `1, or, 0.`

Therefore, let us suppose that, the desired remainder is, `ax+b.`

Symbolically, this can be said, as, let,

`p(x)=(x+2)(x-1)q(x)+(ax+b)..................(ast).`

Now, `p(x),` when divided by `(x+2)` leaves remainder `-19,`

`rArr p(-2)=-19.`

`rArr -2a+b=-19,............[because, (ast)]...........(ast_1).`

On the same lines, `p(1)=2 rArr a+b=2.............(ast_2).`

Solving `(ast_1) and (ast_2), a=7, b=-5,` giving the desired

remainder, `7x-5,` as Respected JIM H. and George C. have

readily obtained.

Enjoy Maths.!